Question

How many conduction electrons are there in a 1.00 mm diameter gold wire that is 40.0 cm long?

Answer 1

Radius of the wire, R = 0.5 mm = 5 * 10^-4 m

Area of the coil, A = $\pi$ R² = 3.14 * (5 * 10^-4)² m² = 7.85 * 10^-7 m²

Volume of the coil, V = A * l = (7.85 * 10^-7 m²) * (0.4 m) = 3.14 * 10^-7 m³

Density of the gold coil, $\rho$ = 19.3 * 10⁶ g m^-3

Mass = Density * Volume

m = (19.3 * 10⁶ g m^-3 ) * (3.14 * 10^-7 m³ ) = 6.0602 g

m = 6.0602 g

Now atomic mass of gold, A = 196.97 g mol^-1

Number of moles,n = (m/A) = (6.0602 / 196.97) mol = 0.0307671219 mol

n = 0.0307671219 mol

Avogadro's number = N_A = 6.022140857 × 10²³ electrons

Number of electrons = n * N_A = 0.0307671219 * (6.022140857 × 10²³ )

N = 1.853 * 10²² electrons